Higher Maths 1.2.3 - Trigonometric Functions

360°

Trigonometric Graphs

Higher Maths 1 2 3 Trigonometric Functions

1

y = sin x

Half of the vertical height.

Amplitude

The horizontal width of one wave section.

Period

Graphs of trigonometric equations are wave shaped with a repeating pattern.

720°

amplitude

period

y = tan x

Graphs of the tangent function:

the amplitudecannot bemeasured.

Amplitude and Period


2

amplitude

y = 3 cos 5 x + 2

For graphs of the form y = a sin bx +

cand

y = a cos bx +

camplitude = a

period = b

360°

a = 3

period= 72°

5360

°

c = 2

Example

For graphs of the form

y = a tan bx +

c

period = b

180°

(amplitude is undefined)

360° degrees = 2 π radians

Radians


3

rr

r60°

rr

r

one radian A radian

is not 60°

Angles are often measuredin radians instead of degrees.A radian is the angle for whichthe length of the arc is the same as the radius.

C = π

DC = 2 π

r

The radius fits into the circumference times.

2 π

r

r

rr

r

r

Radians are normally written as fractions of .

π2 π ≈6.28…Inaccurate

Exact Values of Trigonometric Functions


4

sin x

x0° 30° 45° 60° 90°

not define

d

cos x

tan x

0

1

0

π2

π3

π4

π6

0

21

231

2

13

1 3

23 1

2 21

1

0

4π

Quadrants


5

1st2nd

4th3rd

180°

0°

90°

270°

37°

1st2nd

4th3rd

0

π2

π

23π

π3

4

It is useful to think of angles in terms of quadrants.

37° is in the 1st quadrant

is in the 3rd quadrant

3

Examples

The Quadrant Diagram


6

1st 2nd 4th3rd

all positiv

e

sin positiv

e

tan positiv

e

cos positiv

e

180°

0°

90°

270°

sin +

cos +

tan +

sin +

cos –

tan –

sin –cos

+tan

–

sin –cos

–tan

+

S A

T C

1st2nd

3rd 4th

The Quadrant Diagram

The nature of trigonometric functions can be shown using a simple diagram.

360°

270°

180°

90°

+ +

+ +

7π

T+ C+

A+S+

Quadrants and Exact Values


7

Any angle can be written as an acute angle starting from either 0° or 180°.

S+ A+

T+ C+

120°60°

sin 120°sin 60°

23

=

S+ A+

T+ C+

225°

45°

- cos 45°=12

-

= cos 225°

=

π6

- tan =13

-

tan

=6

-

6-

π6

cos negativ

e

tan negativ

e

7π


8

Solving Trigonometric Equations GraphicallyIt is possible to solve trigonometric equations by sketching a graph.ExampleSolve

2 cos x – 3 = 0for

0 x 2π2 cos x =

23

3

cos x =

x = π6

23√

π 2ππ6

or

x = π62π –

= 11π6

11π6

Sketching y = cos x

gives:

y = cos x

A+

C+

Solving Trigonometric Equations using Quadrants


9

ExampleSolve

2 sin x + 1 = 0

for

0° x 360°12

sin x = -

S+

T+ sin negative

solutions arein the 3rd and 4th quadrants

45°45°

acute angle:

sin -1( )= 45°x = 180° + 45°

= 225°

x = 360° – 45°

= 315°

or

S+ A+

T+ C+

Trigonometric equations can also be solved algebraically using quadrants.

The ‘X-Wing’Diagram

12

3


10

Example 2Solve

tan 4 x + 3 = 0

for

0 x π2

tan 4 x = 3-

A+

C+

S+

T+

tan

negative

solutions arein the 2nd and 4th quadrants

tan -1( ) = π3

4 x = π3π –

= 2π3

x = π6

or 4 x = π32π –

= 5π3

x = 5π12

π

π2

3π2

0

(continued)Solving Trigonometric Equations using Quadrants

acute angle:

Problems involving Compound Angles


11

Solve

6 sin ( 2 x + 10 ) = 3

for

0° x 360°

Example

sin ( 2 x + 10 ) =21

A+

C+

S+

T+

solutions arein the 1st and 2nd quadrants

30°30°

0° x 360°

0° 2 x 720°

10° 2 x + 10 730°

Consider the range: 2 x + 10 = 30° or 150° or 390°

or 510°2 x = 20° or 140° or 380° or 500°x = 10° or 70° or 190° or 250°

360°+30°

360°+150°

Don’t forget to include angles more than 360°


12

( sin x ) 2

sin 2

xis oftenwritten

Solve

7 sin 2

x + 3 sin x – 4 = 0

for

0° x 360°( 7 sin x + 4 ) ( sin x – 1 ) = 0

7 sin x + 4 = 0 sin x – 1 = 0or

sin x = 4- 7sin x = 1S A

T C acute angle ≈ 34.8°

x ≈ 180° + 34.8°

≈ 214.8°

or x ≈ 360° – 34.8°

≈ 325.2°

x = 90°

Example

Solving Quadratic Trigonometric Equations

Higher Maths 1.2.3 - Trigonometric Functions

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